# The ring of SL_2 invariants in sums of conjugation and tautological modules

### Rings of Invariants

Consider $G=SL_2(\mathbb{C})$, and let $V$ be a finite-dimensional $G$-representation. Let $\mathbb{C}[V]$ denote the ring of polynomial functions on the space $V$; it is a free polynomial ring in $\dim(V)$-many variables. Then the ring of invariants of $V$ is the subring $$\mathbb{C}[V]^G:=(f\in\mathbb{C}[V] s.t. \forall g\in G, f(gv)=f(v))$$

### The Conjugation Representation

Let $M_2(\mathbb{C})$ denote the space of $2\times 2$ complex matrices. The conjugation representation $V_C$ of $G$ on $M_2(\mathbb{C})$ is defined by $g\cdot m:= gmg^{-1}$. Thus, $V_C$ is a 4-dimensional $G$-representation.

The ring of invariants $\mathbb{C}[V_C]^G$ is well-known. It is freely generated by the functions $tr$ and $det$; this is a special case of the general fact that the only conjugation-invariant functions on $M_n(\mathbb{C})$ are symmetric functions of eigenvalues.

### The Tautological Module

Let $V_T$ denote the $2$-dimensional $G$-representation where $G$ acts in the 'obvious' way, by the inclusion $G=SL_2(\mathbb{C})\subset GL_2(\mathbb{C})$. This is called the tautological $G$-representation. Since $SL_2(\mathbb{C})$ acts on $V_T$ with a dense orbit, the ring of invariants $\mathbb{C}[V_C]^G$ is boring; it is just $\mathbb{C}$.

Now, let $V_T^n$ denote the $n$-fold direct sum of $V_T$. This is a natural $2n$-dimensional $G$-representation. We can choose a $G$-invariant skew-symmetric bilinear form $\omega$ on $V_T$; this is big words for the usual scalar cross product $v_1\times v_2$. This defines a natural $G$-invariant function on the space of pairs $(v_1,v_2)\in V_T^2$.

Then the ring of invariants $\mathbb{C}[V_T^n]^G$ is generated by $\omega(v_i,v_j)$ for $1\leq i j\leq n$. However, these do not (in general) freely-generate the ring of invariants, there are relations between them. The relations are all of the form $$\omega(v_i,v_k)\omega(v_j,v_l)=\omega(v_i,v_j)\omega(v_k,v_l)+\omega(v_l,v_i)\omega(v_j,v_k)$$ for $1\leq ijkl\leq n$.

Both of these facts can be deduced by observing that $\mathbb{C}[V_T^n]^G$ can be identified with the homogeneous coordinate ring of the Grassmanian $Gr(2,n)$. Then the generators above are the Plucker coordinates, and the relations are the 3-term Plucker relations (there are no higher Plucker relations here).

### The Question

Both of these examples have clever solutions and pretty answers. I am curious about the combination of both cases.

Let $m$ and $n$ be positive integers, and consider the direct sum $G$-representation $V^m_C\oplus V_T^n$. What is the ring of invariants $\mathbb{C}[V_C^m\oplus V_T^n]^G$?

I am aware of general procedures for producing these rings, for arbitrary finite-dimensional $G$-reps; see e.h. Sturmfel's Algorithms in Invariant Theory. However, I suspect that there is a clever solution to this particular problem. Not only because it is a combination of two problems with a clever solution, but because I have a guess as to what the answer is, and all the relations seem to be similar to the 3-term Plucker relations. I also suspect the answer is ancient (like much invariant theory), which is why I am trying to find an answer rather than try to prove my guess is correct by brute force.

• Greg, already the case of simultaneous conjugation of an $m$-tuple of matrices (i.e. $n=0$) isn't particularly easy, albeit classical. The generators can be taken to be traces of the products of length at most 2, but the description of the relations is a bit messy (they follow from the Hamilton-Cayley identity). The classical "symbolic method" tells you how to find generators and relations via polarization. What kind of answer are you hoping for? – Victor Protsak Mar 23 '11 at 2:12
• Victor, I don't know this polarization trick you mean. The solution for multiple conjugation modules is listed in "Sl_2 representations of Finitely Presented Groups" by Brumfiel and Hilden as The First (and Second) Fundamental Theorem(s) of Invariant Theory, and they cite some older works of Procesi. Is this the direction you mean? – Greg Muller Mar 24 '11 at 14:37